X¯¯¯¯ means the class of X mod f(X) in the quotient ring K[X]/(f(x)).
Now, given g∈K[X], consider the associated polynomial function in K[X]/(f(x)). Then, by the definition of the ring operations in K[X]/(f(x)), we have g(u)=g(X¯¯¯¯)=g(X)¯¯¯¯¯¯¯¯¯font-style: inherit; font-variant: inherit; font-weight: inherit; font-stretch: inherit; line-height: no
manpreet
Best Answer
2 years ago
There is a theorem in my book which says the following:
Let KK be a field and let f(X)∈K[X]f(X)∈K[X] be irreducible over KK. Then there exists a field extension L/KL/K, such that ∃u∈L∃u∈L: f(u)=0f(u)=0.
Proof: Notice that K⊆K[X]/(f(x))K⊆K[X]/(f(x)). Clearly, K[X]/(f(x))K[X]/(f(x)) is a field since (f(x))(f(x)) is maximum ideal. If we take u=X¯¯¯¯u=X¯, then f(u)=f(X¯¯